Solution. If we add the equations as they are, neither one of the unknowns will cancel. Now, if the coefficient of y in equation 2) were −4, then the y's would cancel. Therefore we will expand our strategy as follows:

Make one pair of coefficients negatives of one another -- by multiplying both sides of an equation by the same number. Upon adding the equations, that unknown will be eliminated.

To make the coefficients of the y's 4 and −4, we will multiply both sides of equation 2) by 4 :

1)

3x

+

4y

=

19

3x

+

4y

=

19

2)

2x

−

y

=

9

8x

−

4y

=

36

11x

=

55

x

=

5511

x

=

5

The 4 over the arrow in equation 2) signifies that both sides of that equation have been multiplied by 4. Equation 1) has not been changed.

To solve for y, substitute x = 5 in either one of the original equations. In equation 1):

3· 5 + 4y

=

19

4y

=

19 − 15

4y

=

4

y

=

1

The solution is (5, 1).

The student should always verify the solution by replacing x and y with (5, 1) in the original equations.

Example 5. Solve simultaneously:

1)

3x

+

2y

=

−2

2)

2x

+

5y

=

−5

Solution. We must make one pair of coefficients negatives of one another. In this example, we must decide which of the unknowns to eliminate, x or y. In either case, we will make the new coefficients the Lowest Common Multiple (LCM) of the original coefficients -- but with opposite signs.

Thus, if we eliminate x, then we will make the new coeffients 6 and −6. (The LCM of 3 and 2 is 6.) While if we eliminate y, we will make their new coefficients 10 and −10. (The LCM of 2 and 5 is 10.)

The a's are the coefficients of the x's. The b's are the coefficients of the y's. The following is the matrix of those coefficients.

The numbera1b2 − b1a2 is called the determinant of that matrix.

det

=

a1b2 − b1a2

Let us denote that determinant by D.

Now consider this matrix in which the c's replace the coefficients of the x's:

Then the determinant of that matrix -- which we will call Dx -- is

c1b2 − b1c2

And consider this matrix in which the c's replace the coeffients of the y's:

The determinant of that matrix -- Dy -- is

a1c2 − c1a2

Cramer's Rule then states the following:

In every system of two equations in two unknowns in which the determinant D is not 0,

x

=

Dx D

y

=

Dy D

Example. Use Cramer's Rule to solve this system of equations (Problem 7):

5x

+

3y

=

−11

2x

+

4y

=

−10

Solution.

D

=

det

=

5· 4 − 3· 2

=

20 − 6

=

14.

Dx

=

det

=

−11· 4 − 3· −10

=

−44 + 30

=

−14.

Dy

=

det

=

5· −10 − (−11)· 2

=

−50 + 22

=

−28.

Therefore,

x

=

Dx D

=

−14 14

=

−1.

y

=

Dy D

=

−28 14

=

−2.

Problem. Use Cramer's Rule to solve these simultaneous equations.

3x

−

5y

=

−31

2x

+

y

=

1

D

=

det

=

3· 1 − (−5)· 2

=

3 + 10

=

13.

Dx

=

det

=

−31· 1 − (−5)· 1

=

−31 + 5

=

−26.

Dy

=

det

=

3· 1 − (−31)· 2

=

3 + 62

=

65.

Therefore,

x

=

Dx D

=

−26 13

=

−2.

y

=

Dy D

=

6513

=

5.

When the determinant D is not 0, we say that the equations are linearly independent. In any system of linearly independent equations, there is one and only one solution.

When the determinant D is 0, then either 1) there is not a unique solution, it is possible to name many; or 2) there is no solution at all. In case 1), the equations are linearly dependent. One of them is simply a multiple of the other. For example,